C: unsigned char * bytes array copying elements to other unsigned char * array

Question

I think that copying elements of array like this:

unsigned char *new_bytes_array = malloc(sizeof(unsigned char)*length/2); 
for(int i=0, j=0; i<length; i++) {
  if(i % 2 == 0) continue; 
  new_bytes_array[j] = old_bytes_array[i]; 
  j++;

}

is making copy but value not by reference, but I would like to make sure that this will be deep copy not just shallow copy of references or pointers.

I know that this is maybe easy and ridiculous qustion, but I cannot find similar on stack rather memcpy() all array, but I want to copy only some elements for example every second element, or copy 1-3 elements and skip 4th element, (if i%4 == 0 skip element).

Note that sizeof (unsigned char) is always 1, so it's just an annoying way to write 1 (wihch in case of a multiplication is pretty pointless). Consider removing it. — unwind
– unwind, Commented Sep 28, 2016 at 10:59
@unwind actually on different platform, sizeof(unsigned char) is not always 1. So I think it's not a good idea to remove it. — litao3rd
– litao3rd, Commented Sep 28, 2016 at 12:07
@litao3rd No. You are mistaken, that's simply not true. The return value from sizeof is units of char, so by definition sizeof (char) is 1. — unwind
– unwind, Commented Sep 28, 2016 at 12:09
@unwind yep, sizeof(char) is not always 1 on different architecture. And sizeof is executed in compile time, so I think sizeof(unsigned char) is better than a magic number 1. — litao3rd
– litao3rd, Commented Sep 28, 2016 at 12:15

alk · Accepted Answer · 2016-09-28 10:08:48Z

2

new_bytes_array[j] evaluates to an unsigned char.

Assuming old_bytes_array[i] does as well, then this

new_bytes_array[j] = old_bytes_array[i];

copies an unsigned char and not a pointer, not an unsigned char*.

answered Sep 28, 2016 at 10:08

alk

71.2k10 gold badges112 silver badges274 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

unwind · Accepted Answer · 2016-09-28 12:32:22Z

0

Yes, the assignment you show assigns a single unsigned char, i.e. "a byte" on most typical computers, it doesn't copy pointers but raw data.

Here's a complete function that does what you want, and fixes a few minor problems:

void * copy_every_second_byte(const void *src, size_t length)
{
  if(src == NULL || length == 0)
    return NULL;
  unsigned char *out = malloc(length / 2);
  for(size_t i = 0, j = 0; i < length; ++i)
  {
    if(i % 2 == 0) continue;
    out[j++] = ((unsigned char *) src)[i];
  }
  return out;
}

answered Sep 28, 2016 at 12:32

unwind

402k64 gold badges492 silver badges620 bronze badges

Collectives™ on Stack Overflow

C: unsigned char * bytes array copying elements to other unsigned char * array

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related